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 What price glory? (Posted on 2015-01-22)
Two math. wizards play in the following manner:
From a row of integers 0,1,2,…1023,1024 A erases 512 numbers of his choice, - following this B erases 256 numbers of B’s choice.
Step 3: A erases 128 numbers, etc…
So at Step 10 player B chooses one of the 3 remaining numbers and erases it to define the amount of (dollars, pounds, euros, marbles) to be paid by A i.e. the difference between the two remaining numbers.

Clearly, A chooses a strategy to minimize this amount while
his opponent wants to maximize the outcome.

Assuming both follow the best strategy (Which?),
what will be the outcome of the game?

 No Solution Yet Submitted by Ady TZIDON No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 Maybe? | Comment 4 of 5 |
2 is the best answer possible for A.

If we show that A can use a strategy to force an outcome of 2, then 2 will also be the only answer.

Basically the idea is the show that A can always keep the remaining numbers clusterd in one segment after his turn.
if this is true then B will be left with a segment of [a,a+1,a+2]. Obviously B will then choose to erase a+1 and thus the answer will be 2.

So to sum up so far: if A is able to cluster the numbers, this is also the best strategy for him and therefore also the answer.

So:
When it is A's turn, he can erase 2^i numbers. currently on the board (after B deleted 2^(i+1) numbers) are 2^i+1 numbers.

If this is the 3rd turn, A is about to delete 2^7 numbers and there are now 2^8+1 numbers on the board.

Let's look at the previous turn (B's turn), remebering that by induction, there is now only one segment on the board - in any case, the smallest segment he could have left is 2^i - by deleting the numbers in the middle.

Now, in his turn, A can delete this segment (as he can delete 2^i - exactly equal to the smallest possible segment) and we are left with only one segment - As requested.

 Posted by Omri on 2015-01-26 08:03:27

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